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\usepackage{amsmath, amsthm, amssymb, bm, color, framed, graphicx, hyperref, mathrsfs, tasks, enumerate, needspace, tikz-cd, pdfpages}

\title{\textbf{2025微分几何期末复习题}}
\author{唐嘉琪}
\date{\today}
\linespread{1.25  }
\definecolor{shadecolor}{RGB}{241, 241, 255}
\newcounter{problemname}
\newenvironment{problem}{\begin{shaded}\stepcounter{problemname}\par\noindent\textbf{题目\arabic{problemname}. }}{\end{shaded}\par}
\newenvironment{solution}{\par\noindent\textbf{解答. }}{\par}
\newenvironment{note}{\par\noindent\textbf{题目\arabic{problemname}的注记. }}{\par}
\newtheorem{theorem}{定理}
\newtheorem{definition}{定义}
\newtheorem{lemma}{引理}


\begin{document}

\maketitle

\begin{problem}
    给出光滑流形的定义.
\end{problem}

\begin{solution}
    	$M$ 为 $m$ 维流形. $\mathscr A :=\{(U_\alpha,\varphi_\alpha)\}_\alpha$ 给出了 $M$ 上的一簇坐标卡. 称$M$为\textbf{光滑微分流形}, $\mathscr A$ 为 $M$ 的一个 \text{光滑微分结构}, 如果:
	\begin{itemize}
		\item $\mathscr A$ 中的坐标卡覆盖了 $M$;
		\item $\mathscr A$ 中任意两个坐标卡都是光滑相容的;
		\item $\mathscr A$ 是极大的, 即任何与 $\mathscr A$ 中坐标卡均相容的坐标卡都在 $\mathscr A$ 中.
	\end{itemize}
	
		称两个坐标卡 $(U,\varphi_U)$ 与 $(V,\varphi_V)$ 是 光滑相容的, 若 $U\cap V=\varnothing $ 或当 $U\cap V\neq\varnothing $ 相应的坐标变换函数 (具体到 $i$ 个分量)是 $C^\infty$ 连续的.
\end{solution}

\begin{problem}
	证明, 球面$\mathbb S^m$是一个光滑流形.
\end{problem}

\begin{solution}
	$m$维单位球面$\mathbb S^m$定义为
	\begin{align*}
		\mathbb S^m:=\left\{(a_1,\cdots,a_{m+1})\in\mathbb R\colon \sum_{i=1}^{m+1}a_i^2=1\right\}.
	\end{align*}
	
	令$p=(0,\cdots,0,1)$为北极, $q=(0,\cdots,0,-1)$为南极. 开集$U(p)=\mathbb S^m\setminus \{p\}$和$U(q)=\mathbb S^m\setminus \{q\}$之并覆盖$\mathbb S^m$. 下面定义映射$\varphi$ 和$\psi$, 使得$(U(p),\varphi)$和$(U(q),\psi)$是覆盖$\mathbb S^m$的两个$C^\infty$坐标图. 映射$\varphi$和$\psi$由球极投影确定.
	
	对$\forall a\in U(p)$, 记$\lambda$是由点$p$和$a$确定的直线, $\pi$是$\mathbb R^{m+1}$内由$x_{n+1}=0$确定的超平面, $\pi(a)$表示$\mathbb R^{m+1}$内直线$\lambda$和$\pi$相交的点. 当$a=(a_1,\cdots,a_{m+1})$时, 易得$\pi(a)=(x_1,\cdots,x_m,0)$, 这里$x_i=a_i/(1-a_{m+1})$ ($1\leqslant i\leqslant m$). 令
	\begin{align*}
		\varphi(a)=(x_1,\cdots,x_m)=\left(\dfrac{a_1}{1-a_{m+1}},\cdots,\dfrac{a_m}{1-a_{m+1}}\right),
	\end{align*}
	$\varphi$是$U(p)$到$\mathbb R^m$上的一个同胚. 因为如果$(x_1,\cdots,x_m)$已知, 则由计算可得
	\begin{align*}
		a=\left(\dfrac{2x_1}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{2x_2}{1+\sum\limits_{i=1}^mx_i^2},\cdots,\dfrac{2x_m}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{\sum\limits_{i=1}^mx_i^2-1}{1+\sum\limits_{i=1}^mx_i^2}\right).
	\end{align*}
	
	类似的, 定义
	\begin{align*}
		\psi\colon U(q)\to \mathbb R^m,\quad (a_1,\cdots,a_{m+1})\mapsto(y_1,\cdots,y_m),
	\end{align*}
	这里$y_i=a_i/(1+a_{m+1})$. 如果$(y_1,\cdots,y_m)$已知, 则
	\begin{align*}
		a=\left(\dfrac{2y_1}{1+\sum\limits_{i=1}^my_i^2},\dfrac{2y_2}{1+\sum\limits_{i=1}^my_i^2},\cdots,\dfrac{2y_m}{1+\sum\limits_{i=1}^my_i^2},\dfrac{1-\sum\limits_{i=1}^my_i^2}{1+\sum\limits_{i=1}^my_i^2}\right).
	\end{align*}
	由于
	\begin{align*}
		U(p)\cap U(q)=\mathbb S^m\setminus(\{p\}\cup\{q\}),\\
		\forall a\in U(p)\cap U(q),\quad \varphi(a)=(x_1,\cdots,x_m),\\
		\psi(a)=(y_1,\cdots,y_m),
	\end{align*}
	而
	\begin{align*}
		y_i=\dfrac{a_i}{1+a_{m+1}}=\dfrac{x_i}{\sum\limits_{i=1}^mx_i^2},\quad x_i=\dfrac{a_i}{1-a_{m+1}}=\dfrac{y_i}{\sum\limits_{i=1}^my_i^2},
	\end{align*}
	因此$\psi\varphi^{-1}$和$\varphi\psi^{-1}$都是$C^\infty$的, 于是$\mathbb S^m$是一个$m$维光滑流形.\qed
\end{solution}

\begin{problem}
	证明, $f\colon \mathbb S^m\hookrightarrow \mathbb R^{m+1}$为一个嵌入.
\end{problem}

\begin{solution}
	只要证明$f$是一个一一浸入, 且是一个到内的同胚即可.
	
	我们继续沿用上一题的记号, 在球极投影下有(我们选择一个, 另一个是一样的)
	\begin{align*}
		f(x^1,\cdots,x^m)=\left(\dfrac{2x_1}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{2x_2}{1+\sum\limits_{i=1}^mx_i^2},\cdots,\dfrac{2x_m}{1+\sum\limits_{i=1}^mx_i^2},\dfrac{\sum\limits_{i=1}^mx_i^2-1}{1+\sum\limits_{i=1}^mx_i^2}\right).
	\end{align*}
	在不会引起歧义的时候, 为了方便, 后文将$\sum\limits_{i=1}^m$简写为$\sum$.
	
	考察切映射$f_{\star,s}\colon \operatorname T_{\mathbb S^m,s}\to \operatorname T_{\mathbb R^{m+1},f(s)}$的Jacobi阵.
	
	先计算前面$m$个分量的偏导数,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)_{\substack{1\leqslant j\leqslant m\\ 1\leqslant k\leqslant m}}.
	\end{align*}
	当$j=k$时,
	\begin{align*}
		\dfrac{\partial}{\partial x_j}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)=\dfrac{2(1+\sum x_i^2)-2x_j\cdot 2x_j}{(1+\sum x_i^2)^2}=\dfrac{2(1+\sum x_i^2-2x_j^2)}{(1+\sum x_i^2)^2}.
	\end{align*}
	当$j\neq k$时,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{2x_j}{1+\sum x_i^2}\right)=2x_j\dfrac{-2x_k}{(1+\sum x_i^2)^2}=\dfrac{-4x_jx_k}{(1+\sum x_i^2)^2}.
	\end{align*}
	
	再计算第$(m+1)$个分量的偏导数,
	\begin{align*}
		\dfrac{\partial}{\partial x_k}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right)_{1\leqslant k\leqslant m}=\dfrac{2x_k(1+\sum x_i^2)+(1-\sum x_i^2)\cdot 2x_k}{(1+\sum x_i^2)^2}=\dfrac{4x_k}{(1+\sum x_i^2)^2}
	\end{align*}
	
	现在我们来看$\operatorname{Jac}(f_{\star,s})$
	\begin{align*}
		\operatorname{Jac}(f_{\star,s})=\begin{bmatrix}
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_1}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_1}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_1}\left(\dfrac{2x_m}{1+\sum x_i^2}\right)\\[0.4cm]
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_2}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_2}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{2x_2}{1+\sum x_i^2}\right)\\
			\vdots & \vdots & \ddots & \vdots \\
			\dfrac{\partial}{\partial x_1}\left(\dfrac{2x_m}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{2x_m}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{2x_m}{1+\sum x_i^2}\right)\\[0.4cm]
			\dfrac{\partial}{\partial x_1}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right) & \dfrac{\partial}{\partial x_2}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right) & \cdots & \dfrac{\partial}{\partial x_m}\left(\dfrac{\sum x_i^2-1}{1+\sum x_i^2}\right)
		\end{bmatrix}_{(m+1)\times m}\\
		=\dfrac{1}{(1+\sum x_i^2)^2}\begin{bmatrix}
			2(1+\sum x_i^2-2x_1^2) & -4x_1x_2 & \cdots & -4x_1x_m\\
			-4x_2x_1 & 2(1+\sum x_i^2-2x_2^2) & \cdots & -4x_2x_m\\
			\vdots & \vdots & \ddots & \vdots\\
			-4x_mx_1 & -4x_mx_2 & \cdots & 2(1+\sum x_i^2-2x_m^2)\\
			4x_1 & 4x_2 & \cdots & 4x_m
		\end{bmatrix}
	\end{align*}
	我们记上式这个矩阵为$\Lambda$, 即$\operatorname{Jac}(f_{\star,s})=(1+\sum x_i^2)^{-2}\cdot \Lambda$. 对$\Lambda$进行初等变换, 有
	\begin{align*}
		\Lambda\to \begin{bmatrix}
			1 & 0 & \cdots & 0\\
			0 & 1 & \cdots & 0\\
			\vdots & \vdots & \ddots & \vdots \\
			0 & 0 & \cdots & 1\\
			x_1 & x_2 & \cdots & x_m
		\end{bmatrix}
	\end{align*}
	这足以说明$\operatorname{rk}(f_{\star,s})=\operatorname{rk}(\Lambda)=m$, 即$f_{\star,s}$非退化.
	
	$f$的单性是显然的, 这说明了$f$是一个一一浸入, 同时注意到上述球极投影是一个经典的到内同胚, 从而$f\colon \mathbb S^m\hookrightarrow \mathbb R^{m+1}$是一个嵌入.\qed
	
	
	
	%球面$\mathbb S^m$ 在点$s$处的切空间$\operatorname T_{\mathbb S^m, s}$ (在至少同构的意义下) 是$\mathbb R^{m+1}$中与$s$正交的超平面, 即$\operatorname T_{\mathbb S^m, s}\simeq\{v\in\mathbb R^{m+1}\colon s\cdot v=0\}$, 是$\mathbb R^{m+1}$的一个$m$维子空间. 那么切映射
	%\begin{align*}
	%	f_{\star,s}\colon \operatorname T_{\mathbb S^m, s}(\simeq \mathbb R^m)\hookrightarrow \operatorname T_{\mathbb R^{m+1},s}(\simeq \mathbb R^{m+1})
	%\end{align*}
	%是自然的嵌入, 从而是单的. 下图道尽一切.
	%\begin{center}
	%	\begin{tikzcd}
	%		\operatorname{T}_{\mathbb S^m,s}\arrow[d, hookrightarrow, "f_{\star,s}"]\arrow[r, leftrightarrow ,"\sim"] & \mathbb R^m \arrow[d, hookrightarrow, "\text{直接嵌入}g"]\\
	%		\operatorname T_{\mathbb R^{m+1},s}\arrow[r, leftrightarrow, "\sim"]&\mathbb R^{m+1}
	%	\end{tikzcd}
	%\end{center}
\end{solution}

\begin{problem}
	给出外微分算子$\mathrm d$的定义.
\end{problem}

\begin{solution}
	存在唯一的算子$\mathrm d\colon\Gamma(M,\mathcal A^k_M)=\mathcal A^k\to\Gamma(M,\mathcal A^{k+1}_M)=\mathcal A^{k+1}$称为$M$上\textbf{外微分算子}, 若满足:
	\begin{enumerate}
		\item $\mathrm d^2=0$
		\item $f\in C^\infty(M)$, 则$\mathrm df$为$f$的微分, 即$\mathrm df(v)=v(f)$, 对任意$v\in\operatorname T_{M,p}$.
	\end{enumerate}
\end{solution}

\begin{problem}
	证明 Poincáre 引理: 若$\mathrm d$为$M$上外微分算子, 那么$\mathrm d^2 f=\mathrm d^2\omega =0$. 其中$f\in C^\infty(M)$, $\omega\in\mathcal A^k$.
\end{problem}

\begin{solution}
	首先考察$\mathrm d^2$作用在$\Gamma(M,\mathcal A^0_M)=C^{\infty}(M)$上, 对任意$f\in C^\infty(M)$有
	\begin{align*}
		\mathrm d^2f=&\mathrm d(\mathrm df)=\mathrm d\left(\sum_{i=1}^m\dfrac{\partial f}{\partial x^i}\mathrm dx^i\right)=\sum_{i,j=1}^m\dfrac{\partial^2f}{\partial x^j\partial x^i}\mathrm dx^j\wedge\mathrm dx^i\\
		=&\dfrac{1}{2}\sum_{i,j=1}^m\left(\dfrac{\partial^2f}{\partial x^j\partial x^i}-\dfrac{\partial^2f}{\partial x^i\partial x^j}\right)=0
	\end{align*}
	在经一步考察$\mathrm d^2$作用在$\mathcal A^k=\Gamma(M,\mathcal A^k_M)$上之前, 先证明一个引理:
	
	若$\alpha\in\mathcal A^k$, $\beta\in\mathcal A^l$, 则
	\begin{align*}
		\mathrm d(\alpha\wedge\beta)=\mathrm d\alpha\wedge\beta+(-1)^k\alpha\wedge\mathrm d\beta.
	\end{align*}
	
	这个引理的证明也是显然的: 设
	\begin{align*}
		\alpha=&\dfrac{1}{k!}\sum_{i_1,\cdots,i_k=1}^m\alpha_{i_1\,\cdots\,i_k}\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\\
		\beta=&\dfrac{1}{l!}\sum_{j_1,\cdots,j_l=1}^m\beta_{j_1\,\cdots\,j_l}\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}
	\end{align*} 
	那么
	\begin{align*}
		\alpha\wedge\beta=\dfrac{1}{k!l!}\sum_{i_1,\cdots,i_k=1}^m\sum_{j_1,\cdots,j_l=1}^m\alpha_{i_1\,\cdots\,i_k}\beta_{j_1\,\cdots\,j_l}\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\wedge\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}.
	\end{align*}
	\begin{align*}
		\mathrm d(\alpha\wedge\beta)=&\sum_{\substack{1\leqslant i_1<\cdots<i_k\leqslant m\\1\leqslant j_1<\cdots<j_l\leqslant m}}\mathrm d\alpha_{i_1\,\cdots\,i_k}\wedge\beta_{j_1\,\cdots\,j_l}\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\wedge\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}\\
		&+\sum_{\substack{1\leqslant i_1<\cdots<i_k\leqslant m\\1\leqslant j_1<\cdots<j_l\leqslant m}}\alpha_{i_1\,\cdots\,i_k}\mathrm d\beta_{j_1\,\cdots\,j_l}\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\wedge\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}\\
		=&\left(\sum_{1\leqslant i_1<\cdots<i_k\leqslant m}\mathrm d\alpha_{i_1\,\cdots\,i_k}\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\right)\\
		&\wedge\left(\sum_{1\leqslant j_1<\cdots<j_l\leqslant m}\beta_{j_1\,\cdots\,j_l}\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}\right)\\
		&+\left(\sum_{1\leqslant i_1<\cdots<i_k\leqslant m}\alpha_{i_1\,\cdots\,i_k}\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\right)\\
		&\wedge(-1)^k\left(\sum_{1\leqslant j_1<\cdots<j_l\leqslant m}\mathrm d\beta_{j_1\,\cdots\,j_l}\wedge\mathrm dy^{j_1}\wedge\cdots\wedge\mathrm dy^{j_l}\right)\\
		=&\mathrm d\alpha\wedge\beta+(-1)^k\alpha\wedge\mathrm d\beta
	\end{align*}
	现在经一步考察$\mathrm d^2$作用在$\mathcal A^k$上:
	对任意
	\begin{align*}
		\omega=\dfrac{1}{k!}\sum_{i_1,\cdots,i_k=1}^m\omega_{i_1\,\cdots\,i_k}\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\in\mathcal A^k_M.
	\end{align*}
	\begin{equation}\label{wf4}
		\begin{split}
			\mathrm d^2\omega=&\mathrm d(\mathrm d\omega)=\mathrm d\left(\dfrac{1}{k!}\sum_{i_1,\cdots,i_k}^m\mathrm d\omega_{i_1\,\cdots\,i_k}\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\right)\\
			=&\dfrac{1}{k!}\left.\sum_{i_1,\cdots,i_k=1}^m\right(\mathrm d^2\omega_{i_1\,\cdots\,i_k}\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm dx^{i_k}\\
			&+\left.\sum_{\alpha=1}^k(-1)^{\alpha}\mathrm d\omega_{i_1\,\cdots\,i_k}\wedge\mathrm dx^{i_1}\wedge\cdots\wedge\mathrm d^2x^{i_\alpha}\wedge\cdots\wedge\mathrm dx^{i_k}\right)
		\end{split}
	\end{equation}
	注意到$\mathrm d^2\omega_{i_1\,\cdots\,i_k}=\mathrm d^2x^{i_\alpha}=0,\quad\forall\alpha$.那么$\eqref{wf4}=0$.\qed
\end{solution}

\begin{problem}
	证明, $\operatorname T_{M,p}^\star\simeq\mathfrak m_p/\mathfrak m_p^2$.
\end{problem}
\begin{solution}
	认为$M$是$n$维光滑流形. $\forall f\in \mathcal A_{M,p}^0$ 在 $p$ 处使用带有Lagrange余项的 Taylor展开.
	\begin{equation*}
		f(x)=f(p)+\sum_{i = 1}^{n}\left.\frac{\partial f}{\partial x^i}\right|_{p}\left(x^i - p^i\right)+\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left(x^i - p^i\right)\left(x^j - p^j\right).
	\end{equation*}
	其中 $\xi$ 是介于 $x$ 和 $p$ 之间的一个数.\par 
	若 $f\in \mathfrak{m}_{p}$, 则$f(p)=0$.
	此时将 $v_x\in\operatorname T_{M,p} $作用在 $f$ 上, 注意到
	\begin{align*}
		&v_{p}\left[\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left(x^i - p^i\right)\left(x^j - p^j\right)\right]\\
		=&\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left[v_{p}\left(x^i - p^i\right)\left(x^j - p^j\right)+\left(x^i - p^i\right)v_{p}\left(x^j - p^j\right)\right]\\
		=&\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left[0\cdot \left(x^j - p^j\right)+\left(x^i - p^i\right)\cdot 0\right]=0
	\end{align*}
	这样一来
	\begin{equation*}
		v_{p}(f)=\sum_{i=1}^n\left[\left.\dfrac{\partial f}{\partial x^i}\right|_{p}\cdot v_{p}\left(x^i-p^i\right)\right]
	\end{equation*}
	首先应当明确 $x^i-p^i\in \mathfrak m_{p}$.
	因而显然
	\begin{equation*}
		\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left(x^i - p^i\right)\left(x^j - p^j\right)\in \mathfrak m_{p}^2
	\end{equation*}
	设
	\begin{align*}
		g(x)=&\sum_{i = 1}^{n}\left.\frac{\partial f}{\partial x^i}\right|_{p}\left(x^i - p^i\right)\\
		h(x)=&\frac{1}{2}\sum_{i,j = 1}^{n}\left.\frac{\partial^2 f}{\partial x^i\partial x^j}\right|_{\xi}\left(x^i - p^i\right)\left(x^j - p^j\right)
	\end{align*}
	那么 
	\begin{equation*}
		f(x)=g(x)+h(x)
	\end{equation*}
	且 $h(x)\in \mathfrak m_{p}^2$. 这样一来 $f\sim_{\mathfrak m_{p}^2}g$, 记等价类为 $[f]=[g]$. 从而 $\forall f \in \mathfrak m_{p}^2$ 的等价类都可以表示为
	\begin{equation*}
		\sum_{i=1}^n\left.\dfrac{\partial f}{\partial x^i}\right|_{p}\left[x^i-p^i\right]
	\end{equation*}
	从而令 $x\to p$
	\begin{equation*}
		\mathfrak m_{p}\left/\sim_{\mathfrak m_{p}^2}\right. =\mathfrak m_{p}\left/\mathfrak m_{p}^2\right.=\left \langle\left[x^i-p^i\right]_{i=1}^n\right\rangle_{\mathbb R}\simeq \left \langle\left[\left.\mathrm dx^i\right|_{p}\right]_{i=1}^n\right\rangle_{\mathbb R}.
	\end{equation*}
	下面我们来证明$\displaystyle \left \{\left[\left.\mathrm dx^i\right|_{p}\right]\right\}_{i=1}^n$ 是$\displaystyle\operatorname T_{M,p}^\star=\operatorname{Hom}\left(\operatorname T_{M,p},\mathbb R\right)$ 的一组基.
	 
	先回顾一些事实, $x^i$不仅可以作为一个坐标分量(实数), 更是一个函数, 选取局部坐标卡 $\left(U,\varphi=\left(x^1,\cdots,x^n\right)\right)$
	\begin{center}
		\begin{tikzcd}[row sep=0.5em]
			x^i\colon U\arrow[r]&\mathbb R\\
			p\arrow[r,mapsto]&x^i(p)
		\end{tikzcd}
	\end{center}
	这样定义
	\begin{center}
		\begin{tikzcd}[row sep=0.5em]
			\mathrm dx^i\colon\operatorname T_{M,p}\arrow[r]&\mathbb R\\
			v_{p}\arrow[r,mapsto]&v_{p}(x^i)=\displaystyle \sum_{j = 1}^{n}v_{p}^j\left.\dfrac{\partial x^i}{\partial x^j}\right|_{p}=\sum_{j=1}^nv_{p}^j\delta_i^j =v_{p}^i.
		\end{tikzcd}
	\end{center}
	这样简明的证明了
	\begin{equation*}
		\mathfrak m_{p}\left/\mathfrak m_{p}^2\right.=\left \langle\left[x^i-p^i\right]_{i=1}^n\right\rangle_{\mathbb R}\simeq \left \langle\left[\left.\mathrm dx^i\right|_{p}\right]_{i=1}^n\right\rangle_{\mathbb R}=\operatorname T_{M,p}^\star.
	\end{equation*}\qed
\end{solution}

\begin{problem}
	对于任意$\omega\in\mathcal A^1$, 任意$X,Y\in\mathfrak X(M)=\Gamma(M,\operatorname T_M)$, 证明:
	\begin{align*}
		\mathrm d\omega(X,Y)=X(\omega(Y))-Y(\omega X))-\omega ([X,Y]).
	\end{align*}
\end{problem}
\begin{solution}
	设$M$有局部坐标$(M;x^1,\cdots,x^m)$. 在不引起歧义的情况下, 用$\sum\limits_i$代替$\sum\limits_{i=1}^m$. 从而
	\begin{align*}
		X=\sum_i X^i\dfrac{\partial}{\partial x^i},\quad Y=\sum_j Y^j\dfrac{\partial}{\partial x^j},\quad \omega=\sum_k\omega_k\mathrm dx^k.
	\end{align*}
	
	先计算$\omega(Y)$与$\omega(X)$,
	\begin{align*}
		\omega(Y)=&\sum_k\omega_k\mathrm dx^k\left(\sum_jY^j\dfrac{\partial}{\partial x^j}\right)=\sum_k\omega_k\left(\sum_jY^j\delta_j^k\right)=\sum_k\omega_kY^k,\\
		\omega(X)=&\sum_k\omega_k\mathrm dx^k\left(\sum_iX^i\dfrac{\partial}{\partial x^i}\right)=\sum_k\omega_k\left(\sum_iX^i\delta_i^k\right)=\sum_k\omega_kX^k.
	\end{align*}
	
	计算$X(\omega(Y))$与$Y(\omega(X))$,
	\begin{align*}
		X(\omega(Y))=&\sum_iX^i\dfrac{\partial}{\partial x^i}\left(\sum_k\omega_kY^k\right)=\sum_{i,k}X^i\left(\dfrac{\partial\omega_k}{\partial x^i}Y^k+\omega_k\dfrac{\partial Y^k}{\partial x^i}\right),\\
		Y(\omega(X))=&\sum_jY^j\dfrac{\partial}{\partial x^j}\left(\sum_k\omega_kX^k\right)=\sum_{j,k}Y^j\left(\dfrac{\partial\omega_k}{\partial x^j}X^k+\omega_k\dfrac{\partial X^k}{\partial x^j}\right).
	\end{align*}
	
	先计算$[X,Y]$,
	\begin{align*}
		[X,Y]=&\sum_iX^i\dfrac{\partial}{\partial x^i}\left(\sum_jY^j\dfrac{\partial}{\partial x^j}\right)-\sum_jY^j\dfrac{\partial}{\partial x^j}\left(\sum_iX^i\dfrac{\partial}{\partial x^j}\right)\\
		=&\sum_{i,j}X^i\left(\dfrac{\partial Y^j}{\partial x^i}\dfrac{\partial}{\partial x^j}-Y^j\dfrac{\partial^2}{\partial x^i\partial x^j}\right)-\sum_{i,j}Y^j\left(\dfrac{\partial X^i}{\partial x^j}\dfrac{\partial}{\partial x^i} +X^i\dfrac{\partial^2}{\partial x^j\partial x^i}\right)\\
		=&\sum_{i,j}X^i\left(\dfrac{\partial Y^j}{\partial x^i}\dfrac{\partial}{\partial x^j}\right)-\sum_{i,j}Y^j\left(\dfrac{\partial X^i}{\partial x^j}\dfrac{\partial}{\partial x^i}\right).
	\end{align*}
	从而
	\begin{align*}
		\omega([X,Y])=&\sum_k\omega_k\mathrm dx^k\left(\sum_{i,j}\left(X_i\dfrac{\partial Y^j}{\partial x^i}\dfrac{\partial}{\partial x^j}-Y^j\dfrac{\partial X^i}{\partial x^j}\dfrac{\partial}{\partial x^i}\right)\right)\\
		=&\sum_{i,j,k}\omega_k\left(X_i\dfrac{\partial Y^j}{\partial x^i}\delta_j^k-Y^j\dfrac{\partial X^i}{\partial x^j}\delta_i^k\right)\\
		=&\sum_{i,j,k}\omega_k\left(X^i\dfrac{\partial Y^k}{\partial x^i}-Y^j\dfrac{\partial X^k}{\partial x^j}\right).
	\end{align*}
	
	我们整理$X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$有
	\begin{align*}
		&X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\\
		=&\sum_{i,k}X^i\left(\dfrac{\partial\omega_k}{\partial x^i}Y^k+\omega_k\dfrac{\partial Y^k}{\partial x^i}\right)-\sum_{j,k}Y^j\left(\dfrac{\partial\omega_k}{\partial x^j}X^k+\omega_k\dfrac{\partial X^k}{\partial x^j}\right)\\
		&-\sum_{i,j,k}\omega_k\left(X^i\dfrac{\partial Y^k}{\partial x^i}-Y^j\dfrac{\partial X^k}{\partial x^j}\right)\\
		=&\sum_{i,k}X^i\dfrac{
		\partial \omega_k}{\partial x^i}Y^k+\sum_{i,k}X^i\omega_k\dfrac{\partial Y^k}{\partial x^i}-\sum_{j,k}Y^j\dfrac{\partial\omega_k}{\partial x^j}X^k\\
		&-\sum_{j,k}Y^j\omega_k\dfrac{\partial X^k}{\partial x^j}-\sum_{i,k}\omega_kX^i\dfrac{\partial Y^k}{\partial x^i}+\sum_{j,k}\omega_kY^j\dfrac{\partial X^k}{\partial x^j}\\
		=&\sum_{i,k}X^i\dfrac{\partial\omega_k}{\partial x^i}Y^k-\sum_{j,k}Y^j\dfrac{\partial\omega_k}{\partial x^j}X^k.
	\end{align*}
	
	最后来计算$\mathrm d\omega(X,Y)$,
	\begin{align*}
		\mathrm d\omega(X,Y)=&\sum_{k,l}\dfrac{\partial\omega_k}{\partial x^l}\mathrm dx^l\wedge\mathrm dx^k(X,Y)=\sum_{k,l}\dfrac{\partial\omega_k}{\partial x_l}\begin{vmatrix}
			\mathrm dx^l(X) & \mathrm dx^l(Y)\\
			\mathrm dx^k(X) & \mathrm dx^k(Y)
		\end{vmatrix}\\
		=&\sum_{k,l}\dfrac{\partial \omega_k}{\partial x^l}(X^lY^k-X^kY^l)=\sum_{k,l}\left(\dfrac{\partial\omega_k}{\partial x^l}-\dfrac{\partial \omega_l}{\partial x^k}\right)X^lY^k\\
		=&\sum_{i,k}X^i\dfrac{\partial\omega_k}{\partial x^i}Y^k-\sum_{j,k}Y^j\dfrac{\partial\omega_k}{\partial x^j}X^k.
	\end{align*}
	于是命题得证.\qed
\end{solution}
\begin{problem}
	利用 Stokes 公式, 证明:
	\begin{align*}
		\int_{\partial D}P\mathrm dx+Q\mathrm dy=\int_D(Q_x-P_y)\mathrm dx\wedge \mathrm dy.
	\end{align*}
\end{problem}
\begin{solution}
	令$\omega=P\mathrm dx+Q\mathrm dy$, 则
	\begin{align*}
		\mathrm d\omega=&\mathrm d(P\mathrm dx+Q\mathrm dy)=\mathrm dP\wedge\mathrm dx+P\wedge\mathrm d^2x +\mathrm dQ\wedge\mathrm dy+Q\mathrm d^2y\\
		=&\left(P_x\mathrm dx+P_y\mathrm dy\right)\wedge\mathrm dx+\left(Q_x\mathrm dx+Q_y\mathrm dy\right)\wedge\mathrm dy\\
		=&P_x\mathrm d^2x+P_y\mathrm dy\wedge\mathrm dx+Q_x\mathrm dx\wedge\mathrm dy+Q_y\mathrm d^2y=(Q_x-P_y)\mathrm dx\wedge \mathrm dy
	\end{align*}
	代入$\int_{\partial D}\omega=\int_{D}\mathrm d\omega$得证.\qed
\end{solution}

\begin{problem}
	设$\varphi\colon M\to N$, $\psi\colon N\to X$为光滑流形$M,N,X$之间的光滑映射, 给出$\varphi_\star, \psi_\star, \varphi^\star, \psi^\star$的定义, 并证明:
	\begin{align*}
		(\psi\circ\varphi)_\star=\psi_\star\circ\varphi_\star,\quad(\psi\circ\varphi)^\star=\varphi^\star\circ\psi^\star.
	\end{align*}
\end{problem}
\begin{solution}
	$\varphi_{\star,p}\colon \operatorname T_{M,p}\to \operatorname T_{N,\varphi(p)},\quad v\mapsto\varphi_{\star,p}(v)$, 满足对任意$f\in C^\infty(N)$, 有
	\begin{align*}
		\varphi_{\star,p}(v)(f)=v(f\circ \varphi).
	\end{align*}
	
	$\varphi_p^\star\colon \operatorname T_{N,\varphi(p)}^\star\to \operatorname T_{M,p}^\star,\quad \omega\mapsto\varphi_p^\star(\omega)$, 满足对任意$v\in\operatorname T_{M,p}$, 有
	\begin{align*}
		\varphi_p^\star(\omega)(v)=\omega(\varphi_{\star,p}(v))
	\end{align*}
	
	对于$\psi_{\star,\varphi(p)}$与$\psi_{\varphi(p)}^\star$的定义是类似的.
	
	我们约定
	\begin{center}
		\begin{tikzcd}[row sep=0.5em]
			M\arrow[r,"\varphi"] & N\arrow[r,"\psi"] & X\\
			m\arrow[r,mapsto] & n=\varphi(m)\arrow[r,mapsto] & x=(\psi\circ \varphi)(m)
		\end{tikzcd}
	\end{center}
	对于任意$v\in \operatorname T_{M,m}$, $f\in C^\infty(X)$, 有
	\begin{align*}
		(\psi\circ\varphi)_{\star,m}(v)(f)=v(f\circ\psi\circ\varphi)=(\varphi_{\star,m}(v))(f\circ \psi)=\psi_{\star,n}(\varphi_{\star,m}(v))(f).
	\end{align*}
	对任意$\omega\in \operatorname T_{X,x}^\star$, $v\in \operatorname T_{M,m}$, 有
	\begin{align*}
		(\psi\circ\varphi)_m^\star(\omega)(v)=\omega((\psi\circ\varphi)_{\star,m}(v))=\omega(\psi_{\star,n}(\varphi_{\star,m}(v)))\\
		=\psi_n^\star(\omega)(\varphi_{\star,m}(v))=\varphi_m^\star(\psi_n^\star(\omega))(v)
	\end{align*}
	从而我们证明了结论$(\psi\circ\varphi)_{\star,m}=\psi_{\star,n}\circ\varphi_{\star,m},\quad (\psi\circ\varphi)_m^\star=\varphi_m^\star\circ\psi_n^\star$.\qed
\end{solution}

\begin{problem}
	给出同伦的定义, 并证明: 
	\begin{itemize}
		\item 若流形$M$与$N$是同伦等价的, 那么其上同调群是同构的.
		\item 若$M$是同伦可缩的, 则其de Rham上同调群$H_{\operatorname{dR}}^k(M,\mathbb R)=\begin{cases}
			\{0\},& k\geqslant 1,\\
			\mathbb R,& k=0.
		\end{cases}$.
	\end{itemize}
\end{problem}
\begin{solution}
	设$X$, $Y$为拓扑空间, $f,g\colon X\to Y$为两个连续映射. 若存在连续映射$H\colon X\times [0,1]\to Y$使得对于任何$x\in X$满足:
	\begin{enumerate}
		\item $H(x,0)=f(x)$,
		\item $H(x,1)=g(x)$.
	\end{enumerate}
	则称$f$与$g$是\textbf{同伦的(homotopic)}. 映射$H$被称为从$f$到$g$的\textbf{同伦(homotopy)}.
	
	若存在光滑映射$\varphi\colon M\to N$与$\psi\colon N\to M$使得$\varphi\circ\psi$同伦于$\operatorname{id}_N$, 且$\psi\circ\varphi$同伦于$\operatorname{id}_M$, 则称这两个流形$M$, $N$是\textbf{同伦等价}的.
	
	为了证明de Rham上同调群的同伦不变性, 先证明函子同伦不变性: 
	\begin{theorem}[函子同伦不变性]
	设$f,g\in C^\infty(M,N)$是同伦的, 则\begin{align*}
		f^\star=g^\star\colon H_{\operatorname{dR}}^k(N,\mathbb R)\to H_{\operatorname{dR}}^k(M,\mathbb R).
	\end{align*}
	\end{theorem}
	证明是构造如下图所示的上链同伦
	\begin{center}
		\begin{tikzcd}
			\cdots\arrow[r,"\mathrm d"] &\mathcal A^{k-1}(N)\arrow[r,"\mathrm d"]\arrow[ld, dashed, swap, "h_{k-1}"]\arrow[d, "f^\star", "g^\star"'] &\mathcal A^k(N)\arrow[r,"\mathrm d"]\arrow[ld, dashed, swap, "h_k"]\arrow[d, "f^\star", "g^\star"'] &\mathcal A^{k+1}(N)\arrow[r,"\mathrm d"]\arrow[ld, dashed, swap, "h_{k+1}"]\arrow[d, "f^\star", "g^\star"'] &\cdots\arrow[ld, dashed, swap, "h_{k+2}"] \\
			\cdots\arrow[r,"\mathrm d"] &\mathcal A^{k-1}(M)\arrow[r,"\mathrm d"] &\mathcal A^k(M)\arrow[r,"\mathrm d"] &\mathcal A^{k+1}(M)\arrow[r,"\mathrm d"] &\cdots 
		\end{tikzcd}
	\end{center}
	先假设上链同伦存在, 我们证明函子同伦不变性. 对任意$[\omega]_{\operatorname dR}\in H_{\operatorname{dR}}^k(N,\mathbb R)$, 有
	\begin{align*}
		g^\star\omega-f^\star\omega=(\mathrm dh+h\mathrm d)\omega=\mathrm d(h\omega)\in B^k(M,\mathbb R).
	\end{align*}
	因此$f^\star([\omega]_{\operatorname{dR}})=[f^\star\omega]_{\operatorname{dR}}=[g^\star\omega]_{\operatorname{dR}}=g^\star([\omega]_{\operatorname{dR}})$.
	
	最后构造上链同伦, 工具是使用特定向量场生成的流, 先证明一个引理
	\begin{lemma}
		设$X$为$M$上完备向量场, $X_t$为$X$生成的流, 则存在线性算子$Q_k\colon \mathcal A^k(M)\to \mathcal A^{k-1}(M)$使得对任意的$\omega\in \mathcal A^k(M)$有
		\begin{align*}
			X_1^\star\omega-\omega=\mathrm dQ_k(\omega)+Q_{k+1}(\mathrm d\omega).
		\end{align*}
	\end{lemma}
	首先, 直接计算可得
	\begin{align*}
		\dfrac{\mathrm d}{\mathrm dt}X_t^\star\omega=\left.\dfrac{\mathrm d}{\mathrm ds}\right|_{s=0}X_{t+s}^\star\omega=\left.\dfrac{\mathrm d}{\mathrm ds}\right|_{s=0}X_s^\star X_t^\star \omega=\mathcal L_{X}(X_t^\star\omega)=\mathrm di_X(X_t^\star\omega)+i_X\mathrm d(X_t^\star\omega),
	\end{align*}
	于是对于$\omega\in\mathcal A^k(M)$, 若令$Q_k(\omega)=\int_{0}^1i_X(X_t^\star\omega)\mathrm dt$, 则$Q_k\colon \mathcal A^k(M)\to \mathcal A^{k-1}(M)$, 且
	\begin{align*}
			X_1^\star\omega-\omega=\int_0^1\left(\dfrac{\mathrm d}{\mathrm dt}X_t^\star\omega\right)\mathrm dt =\mathrm dQ_k(\omega)+Q_{k+1}(\mathrm d\omega).
	\end{align*}
	现在利用该引理完成函子同伦不变性证明的最后一步, 即构造上链同伦$h_k\colon \mathcal A^k(N)\to \mathcal A^{k-1}(M)$: 设$W=M\times \mathbb R$, 则$X=\dfrac{\partial}{\partial t}$是$W$上的完备向量场, 且它生成的流是
	\begin{align*}
		X_t(p,a)=(p,a+t).
	\end{align*}
	由引理, 存在线性算子$Q_k\colon \mathcal A^k(W)\to \mathcal A^{k-1}(W)$使得
	\begin{align*}
		X_1^\star\omega-\omega=\mathrm dQ_k(\omega)+Q_{k+1}(\mathrm d\omega).
	\end{align*}
	在根据流形上的反函数定理, 任意两个同伦的光滑映射都是光滑同伦. 设$F\colon W\to N$是$f$和$g$之间的光滑同伦, 并设$\iota\colon M\hookrightarrow W$是包含映射$\iota(p)=(p,0)$, 则
	\begin{align*}
		f=F\circ\iota\quad\text{且}\quad g=F\circ X_1\circ\iota,
	\end{align*}
	由此对任意$\omega\in\mathcal A^k(N)$有
	\begin{align*}
		g^\star\omega-f^\star\omega=\iota^\star X_1^\star F^\star\omega-\iota^\star F^\star\omega=\iota^\star(\mathrm dQ_k+Q_{k+1}\mathrm d)F^\star\omega=(\mathrm d\iota^\star Q_kF^\star+\iota^\star Q_{k+1}F^\star\mathrm d)\omega.
	\end{align*}
	所以如果记$h_k=\iota^\star Q_kF^\star$, 则$h_k\colon \mathcal A^k(M)\to \mathcal A^{k-1}(N)$满足
	\begin{align*}
		g^\star\omega-f^\star\omega=(\mathrm dh_k+h_{k+1}\mathrm d)\omega,
	\end{align*}
	从而这些$h_k$就是所需的上链同伦.
	
	到这里已经成功证明了定理(函子同伦不变性), 现用其证明de Rham上同调群的同伦不变性: 设$\varphi\colon M\to N$和$\psi\colon N\to M$是连续映射且满足$\varphi\circ\psi\sim\operatorname{id}_N$和$\psi\circ\varphi\sim\operatorname{id}_M$. 根据连续映射的Whitney逼近定理, 光滑流形之间的任意连续映射都与某个光滑映射同伦, 故存在$\varphi_1\in C^\infty(M,N)$和$\psi_1\in C^\infty(M,N)$使得$\varphi_1\sim\varphi$且$\psi_1\sim\psi$. 因此$\varphi_1\circ \psi_1$和$\psi_1\circ\varphi_1$都是光滑映射, 而且$\varphi_1\circ \psi_1\sim\operatorname{id}_N$, $\psi_1\circ\varphi_1\sim\operatorname{id}_M$. 于是利用函子性并应用函子同伦不变性可得
	\begin{align*}
		\varphi_1^\star\circ\psi_1^\star=&\operatorname{id}\colon H_{\operatorname{dR}}^k(M,\mathbb R)\to H_{\operatorname{dR}}^k(M,\mathbb R)\\
		\psi_1^\star\circ\varphi_1^\star=&\operatorname{id}\colon H_{\operatorname{dR}}^k(N,\mathbb R)\to H_{\operatorname{dR}}^k(N,\mathbb R)
	\end{align*}
	所以$\varphi^\star$和$\psi^\star$是线性同构.
	
	到这里我们证明了de Rham上同调群的同伦不变性, 而对于题目后半部分``若$M$是同伦可缩的, 则其de Rham上同调群$H_{\operatorname{dR}}^k(M,\mathbb R)=\{0\},k\geqslant 1;=\mathbb R, k=0$.''则是de Rham上同调群的同伦不变性的直接推论.\qed
\end{solution}
\begin{note}
	\begin{definition}[上链同伦]
		设$f,g\in C^\infty(M,N)$是同伦的. 若是一列映射$h_k\colon\mathcal A^k(N)\to \mathcal A^{k+1}(M)$满足
		\begin{align*}
			g^\star-f^\star=\mathrm d_Mh_k+h_{k+1}\mathrm d_N.
		\end{align*}
		则称映射列$h=(h_k)$是$f^\star$和$g^\star$之间的一个\textbf{上链同伦}.
	\end{definition}
	\begin{theorem}[连续函数的Whitney逼近定理]
		设$M$, $N$是光滑流形, $g\in C^0(M,N)$是连续映射. 则存在同伦于$g$的光滑映射$f\in C^\infty(M,N)$. 此外, 若$g$在闭子集$A\subset M$上光滑, 则可以使得$f|_A=g|_A$.
	\end{theorem}
\end{note}

\begin{problem}
	给出Lie群, Lie代数的定义, 并证明:
	\begin{itemize}
		\item 向量加法群$\mathbb R$是一个Lie群, 其上典范的Lie代数为$1$维平凡李代数$\mathfrak g=\mathbb R$.
		\item 一般线性群$\operatorname{GL}(n,\mathbb R)$为一个$n^2$维非紧Lie群, 其上典范Lie代数是$\mathfrak{gl}(n,\mathbb R)=\operatorname M_{n\times n}(\mathbb R)$.
	\end{itemize}
\end{problem}

\begin{solution}
	设$G$是一个群, 如果$G$上还赋有一个光滑流形结构, 满足映射$G\times G\to G,\quad (g_1,g_2)\mapsto g_1\cdot g_2^{-1}$是光滑的, 则称$G$为一个\textbf{Lie群}.
	
	$V$是$\mathbb R$-向量空间(可以在更一般的域中), 上面有一个二元括号运算
	$\lbrack\cdot,\cdot\rbrack\colon V\times V\to V$, 
	使得对任意的$u,v,w\in V$, 以及$a,b\in \mathbb R$, 都有
	\begin{enumerate}
		\item (反对称性)$[u,v]=-[v,u]$,
		\item (线性性)$[au+bv,w]=a[u,w]+b[v,w]$,
		\item (Jacobi恒等式)$[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0$,
	\end{enumerate}
	则称$(V,\lbrack\cdot,\cdot\rbrack)$是一个\textbf{Lie代数(Lie algebra)}$\lbrack\cdot,\cdot\rbrack$为该Lie代数的\textbf{Lie括号}.
	
	Lie代数$\mathfrak g:=\{G
	,\text{全体左不变向量场}\}$被称为\textbf{Lie群$G$的(典范的)Lie代数}.
	
	现在来分别证明命题:
	
	显然$\mathbb R^1$在向量加法运算下是一个Lie群, 这是因为映射$(x,y)\mapsto x-y$是光滑的. 不仅如此, 对任意$a\in\mathbb R$, 左平移$L_a$是$\mathbb R^1$上的平移映射, 故左不变向量场实际上就是常值向量场
	\begin{align*}
		X=\lambda\dfrac{\partial}{\partial x^1},\quad\lambda\in \mathbb R.
	\end{align*}
	注意到$\mathfrak g=\left\{\lambda\dfrac{\partial}{\partial x^1}\colon\lambda\in\mathbb R\right\}$与$\mathbb R^1$在Lie括号下完全一致, 自然有同构$\mathfrak g\simeq \mathbb R^1$, 又有任意两个左不变向量场的Lie括号为$0$, 对应为任意两个$R^1$中的元素Lie括号为$0$. 换而言之, 向量加法群$\mathbb R^1$的Lie代数就是$1$维平凡Lie代数$\mathbb R^1$.
	
	$\operatorname{GL}(n,\mathbb R)$是一个$n^2$维的非紧Lie群, 它是不连通的, 而且恰有两个连通分支, 即
	\begin{align*}
		\operatorname{GL}_+(n,\mathbb R)=&\{X\in\operatorname M_{n\times n}(\mathbb R)\colon \operatorname X>0\},\\
		\operatorname{GL}_-(n,\mathbb R)=&\{X\in\operatorname M_{n\times n}(\mathbb R)\colon \operatorname X<0\}.
	\end{align*}
	由于$\operatorname{GL}(n,\mathbb R)$是$\operatorname{M}_{n\times n}\simeq \mathbb R^{n^2}$的一个开子集, 所以$\operatorname{GL}(n,\mathbb R)$的Lie代数的底空间(即$\operatorname{GL}(n,\mathbb R)$在$e=I_n$处的切空间), 就是集合$\operatorname M_{n\times n}(n,\mathbb R)$自身,
	\begin{align*}
		\mathfrak{gl}(n,\mathbb R)=\operatorname M_{n\times n}(\mathbb R).
	\end{align*}
	为了求出Lie括号运算, 任取矩阵$A=(A_ij)_{n\times n}\in\mathfrak{gl}(n,\mathbb R)$, 并记$\operatorname{GL}(n,\mathbb R)$上的全局坐标系维$(X^{ij})$. 那么跟$A$相应的在$\operatorname T_{\operatorname{GL}(n,\mathbb R),I_n}$处向切的切向量是$\sum A_{ij}\dfrac{\partial}{\partial X^{ij}}$, 而它生成的左不变向量场在矩阵$X=(X^{ij})$处的向量为$\sum X^{ik}A_{kj}\dfrac{\partial}{\partial X^{ij}}$. 从而矩阵$A,B\in \mathfrak g$之间的Lie括号$[A,B]$是对应于向量场
	\begin{align*}
		\left[\sum X^{ik}A_{kj}\dfrac{\partial}{\partial X^{ij}},\sum X^{pq}B_{qr}\dfrac{\partial}{\partial X^{pr}}\right]=&\sum X^{ik}A_{kj}B_{jr}\dfrac{\partial}{\partial X^{ir}}-\sum X^{pq}B_{qr}A_{rj}\dfrac{\partial}{\partial X^pj}\\
		=&\sum X^{ik}(A{kr}B_{rj}-B_{kr}A_{rj})\dfrac{\partial}{\partial X^{ij}}.
	\end{align*}
	的矩阵. 换而言之, $\mathfrak g$上的Lie括号就是矩阵换位子$[A,B]=AB-BA$.
	
\end{solution}

\begin{problem}
	对于Lie群$G$上的一个形式$\omega$, 如果对每一个$\sigma\in G$, 都有$L_\sigma^\star\omega=\omega$, 则称之为左不变的. $E_{l\operatorname{inv}}^p(G)$表示$G$上左不变$p$形式构成的向量空间. 令
	\begin{align*}
		E_{l\operatorname{inv}}^\star(G)=\bigoplus_{p=0}^{\dim G}E_{l\operatorname{inv}}^p(G).
	\end{align*}
	证明:
	\begin{itemize}
		\item 左不变形式是光滑的.
		\item $E_{l\operatorname{inv}}^\star(G)$是由$G$上的所有光滑形式构成的代数$E^\star(G)$的一个子代数. 而且映射$\omega\mapsto\omega(e)$是从$E_{l\operatorname{inv}}^\star(G)$到$\Lambda(G_e^\star)$上的一个代数同构, 其中$\Lambda(G_e^\star)$为$G_e^\star$的外代数. 特别的, 这个映射给出$E_{l\operatorname{inv}}^1(G)$到$G_e^\star$, 因而也是$\mathfrak g^\star$的一个自然同构. 以这种方式把$E_{l\operatorname{inv}}^1(G)$看做$G$的Lie代数的对偶空间.
		\item 如果$\omega$是一个左不变的$1$形式而$X$是一个左不变的向量场, 那么$\omega(X)$是$G$上的一个常函数.
		\item 如果$\omega\in E_{l\operatorname{inv}}^1(G)$且$X,Y\in\mathfrak g$, 那么$\mathrm d\omega(X,Y)=-\omega[X,Y]$.
		\item 令$\{X_1,\cdots,X_d\}$是$\mathfrak g$的一个基, 并且以$\{\omega_1,\cdots,\omega_d\}$作为$E_{l\operatorname{inv}}^1(G)$的对偶基, 那么存在$G$关于$\mathfrak g$的基$\{X_i\}_{i=1}^d$的结构常数$c_{ij}^k$使得$[X_i,X_j]=\sum\limits_{k=1}^dc_{ij}^kX_k$, 且满足\begin{align*}
			\begin{cases}
				c_{ij}^k+c_{ji}^k=0,\\
				\sum\limits_r(c_{ij}^rc_{rk}^s+c_{jk}^rc_{ri}^s+c_{ki}^rc_{rj}^s)=0.
			\end{cases}
		\end{align*}$\omega_i$的外导数由 Maurer-Cartan 方程$\mathrm d\omega_i=\sum\limits_{j<k}c_{jk}^i\omega_k\wedge\omega_j$给出.
	\end{itemize}
\end{problem}

\begin{solution}
	对1,3,4,5点作证明.
	
	\begin{itemize}
		\item 若$\omega\in \mathcal A^r(G)$. 只需证明对$G$上所有光滑$X_i\in\mathfrak X(G), (1\leqslant i\leqslant r)$有$\omega (X_1,\cdots,X_r)$是光滑的, 注意到有
		\begin{align}\label{12-1}
			\omega(X_1,\cdots,X_r)(\sigma)=&\omega_\sigma(X_1|_\sigma,\cdots,X_r|_\sigma)=(L_{\sigma^{-1}}^\star\omega_e)(X_1|_\sigma,\cdots,X_r|_\sigma)\notag \\
			=&\omega_e(L_{\star,{\sigma^{-1}}}X_1|_\sigma,\cdots,L_{\star,{\sigma^{-1}}}X_r|_\sigma)
		\end{align}
		所以只需验证\eqref{12-1}是光滑的.
		
		另设$\widetilde \omega\in\mathcal A^r(G)$满足$\widetilde\omega_e=\omega_e$. 对于$\varphi\colon G\times G,(a,b)\mapsto ab.$, 有$\varphi^\star \widetilde\omega$是光滑的. 此外, $[0,X_i]\in\mathfrak X(G\times G)$, 且$\iota\colon G\to G\times G, \tau\mapsto(\tau^{-1},\tau)$是光滑的. 因此 $$\varphi^\star\widetilde\omega([0,X_1],\cdots,[0,X_r])(\iota(\sigma))$$关于$\sigma$是光滑的. 但
		\begin{align}\label{12-2}
			\varphi^\star\widetilde\omega([0,X_1],\cdots,[0,X_r])(\iota(\sigma))=&\varphi^\star\widetilde\omega_{(\sigma^{-1},\sigma)}([0,X_1]_{(\sigma^{-1},\sigma)},\cdots,[0,X_r]_{(\sigma^{-1},\sigma)})\notag \\
			=&\widetilde\omega_e(\varphi_\star[0,X_1]_{(\sigma^{-1},\sigma)},\cdots,\varphi_\star[0,X_r]_{(\sigma^{-1},\sigma)})
		\end{align}
		
		此外,
		\begin{align*}
			\varphi_\star[0,X_i]_{(a,b)}(f)=&[0,X_i]_{(a,b)}(f\circ \varphi)=0|_a(f\circ \varphi \circ \iota_b^1)+X_i|_b(f\circ \varphi\circ \iota b^2)\\
			=&X_i|_b(f\circ L_a)=L_{\star,a}X_i|_b(f)
		\end{align*}
		从而$\eqref{12-2}=\omega_e(L_{\star,\sigma^{-1}}X_1|_\sigma,\cdots,L_{\star,\sigma^{-1}}X_r|_\sigma)$.
		\item $\omega(X)(\sigma)=\omega_\sigma(X|_\sigma)=(L_{\sigma^{-1}}^\star\omega)(X|_\sigma)=\omega_e(L_{\star,\sigma^{-1}}X|_\sigma)=\omega_e(X|_e)=\omega(X)(e)$.
		\item $\mathrm d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega[X,Y]=-\omega[X,Y]$.
		\item $\mathfrak g$在Lie括号下封闭$\Longrightarrow$ $c_{ij}^k$的存在性.
		
		由Lie括号性质$\Longrightarrow$ $\begin{cases}
				c_{ij}^k+c_{ji}^k=0,\\
				\sum\limits_r(c_{ij}^rc_{rk}^s+c_{jk}^rc_{ri}^s+c_{ki}^rc_{rj}^s)=0.
			\end{cases}$.
			
		由于$L_{\sigma}^\star$与$\mathrm d$可交换, $\mathrm d\omega_i\in E_{l\operatorname{inv}}^2(G)$, 且$\mathrm d\omega_i(X_j,X_k)=-\omega[X_j,X_k]$, 推得
		\begin{align*}
			\mathrm d\omega_i=\sum\limits_{j<k}c_{jk}^i\omega_k\wedge\omega_j+\omega^\prime.
		\end{align*}
		其中$\omega^\prime(X,Y)=0,\forall X,Y\in\mathfrak g$, 但由于$E_{l\operatorname{inv}}^2(G)\stackrel{\sim}{\longrightarrow}\Lambda^2(G_e)\simeq\mathfrak g\wedge \mathfrak g$为同构, 从而$\omega^\prime=0$.
	\end{itemize}
\end{solution}

\begin{problem}
	计算一般线性群$\operatorname{GL}(n,\mathbb R)$的结构常数.
\end{problem}

\begin{solution}
	$\dim \operatorname{GL(n,\mathbb R)}=n^2$, 其上有标准基$\{E_{ij}\}_{\substack{1\leqslant i\leqslant n\\1\leqslant j\leqslant n}}$, 其中$E_{ij}$表示第$(i,j)$元为$1$, 其余为$0$的矩阵. 我们注意到$E_{ij}E_{kl}=\delta_j^kE_{il}$, 这样以来,
	\begin{align*}
		[E_{ij},E_{kl}]=&E_{ij}E_{kl}-E_{kl}E_{ij}=\delta_j^kE_{il}-\delta_l^iE_{kj}\\
		=&\sum_{m,n}\delta_j^k\delta_i^m\delta_l^nE_{mn}-\sum_{m,n}\delta_l^i\delta_k^m\delta_j^nE_{mn}.
	\end{align*}
	由于结构常数满足$[E_{ij},E_{kl}]=\sum\limits_{m,n}c_{(ij)(kl)}^{(mn)}E_{mn}$. 比较两式便得
	\begin{align*}
		c_{(ij)(kl)}^{(mn)}=\delta_j^k\delta_i^m\delta_l^n-\delta_l^i\delta_k^m\delta_j^n.
	\end{align*}
\end{solution}

\end{document}
